Question

4 solid cubes were made out of the same material. Each of the four side cubes have different side lengths which are: 6cm, 8cm, 10cm, and 12cm. Your task is to divide the cubes onto two sides of a scale so that the two sides are even. Which cube (or cubes) do you put on one side, and subsequently which remaining cube (or cubes) do you put on the other side?

Answer 1

To distribute four solid cubes with side lengths of 6cm, 8cm, 10cm, and 12cm across a scale evenly, we must calculate their individual volumes. Combining the volumes of the 6cm and 8cm cubes does not equal the sum of the 10cm and 12cm cubes, meaning it's not possible to balance the scale with intact cubes.

Step-by-step explanation:

The question involves balancing two sides of a scale by finding combinations of cubes with different side lengths that have equal volumes. The cubes have side lengths of 6cm, 8cm, 10cm, and 12cm. To solve this problem, one must calculate the volume of each cube by raising the side length to the power of three (since the volume of a cube is equal to the side length cubed). The cubes with side lengths of 6cm and 8cm (216 cubic cm and 512 cubic cm, respectively) can be balanced against the cubes with side lengths of 10cm and 12cm (1000 cubic cm and 1728 cubic cm, respectively), by finding two pairs that sum up to the same total volume.

Using this approach:

• Cube with 8cm side: 8cm x 8cm x 8cm = 512 cm³
• Cube with 6cm side: 6cm x 6cm x 6cm = 216 cm³
• Cube with 10cm side: 10cm x 10cm x 10cm = 1000 cm³
• Cube with 12cm side: 12cm x 12cm x 12cm = 1728 cm³

By combining the 8cm and 6cm cubes (512 cm³ + 216 cm³ = 728 cm³) on one side and the 10cm cube (1000 cm³) on the other side, we get a balanced scale, with a discrepancy of 272 cm³. To address this, we subtract a cube of 12cm, portioning out a section of its volume to the other side. However, to perfectly balance the scale the remaining cubes' volumes would need to be split, which is not possible as cube volume cannot be partially allocated without altering its integrity. Therefore, this problem does not have a solution where intact cubes can be used to balance the scale.

Answer 2

Since all cubes are made of the same material, they have the same density. This means that equal pieces taken from each of the cubes will have the same mass, and therefore weight. This means that all we need to know is volume- how big the cubes are.

Volume is three dimensional, so we need to use units cubed. This is helpful since with actual cubes, you can just multiply the side length by itself three times.

The volumes are:

• 6 cm: 216 cm^3 (6*6*6)
• 8 cm: 512 cm^3 (8*8*8)
• 10 cm: 1000 cm^3 (10*10*10)
• 12 cm: 1728 cm^3 (12*12*12)

Again, because the densities are the same we only need the volume. If they weren't we would have to calculate the mass.

The final step is just to figure out how to balence the equation, and therefore the scale. The volume on side A needs to equal the volume on side B. Here, the volumes of the three smallest cubes equals the volume of the largest.

216+512+1000=1728

The reason the differences between the cubes gets bigger each time, despite adding exactly 2 cm, is that the values are cubed. The difference between x and y is much smaller than the difference between x^n and y^n.